A bullet is fired vertically upward with speed v from the surface of a spherical planet. At its maximum height, the gravitational acceleration is one-fourth of its surface value. If the planet's escape speed is V e s c V_{esc} V esc = v√N, find N.
Physics · Gravitation · Acceleration Due to Gravity & Escape Velocity
JEE Advanced 2015

Question

A bullet is fired vertically upward with speed v from the surface of a spherical planet. At its maximum height, the gravitational acceleration is one-fourth of its surface value. If the planet's escape speed is V e s c V_{esc} V esc = v√N, find N.

Physics · Gravitation · Acceleration Due to Gravity & Escape Velocity

JEE Advanced 2015

G1 Physics · Accepted Answer

Since g ∝ 1/r² outside the planet, g at maximum height being g₀/4 means rₘₐₓ = 2R. Energy conservation givₑ (1/2)v² - GM/R = -GM/(2R), hence v² = GM/R. Escape speed satisfies vₑₛc² = 2GM/R = 2v². Therefore vₑₛc = v√2 and N = 2.

A bullet is fired vertically upward with speed v from the surface of a spherical planet. At its maximum height, the gravitational acceleration is one-fourth of its surface value. If the planet's escape speed is V e s c V_{esc} V esc ​ = v√N, find N.

A bullet is fired vertically upward with speed v from the surface of a spherical planet. At its maximum height, the gravitational acceleration is one-fourth of its surface value. If the planet's escape speed is V e s c V_{esc} V esc = v√N, find N.